Generalized Bent Functions and Their Gray Images

Abstract

In this paper we prove that generalized bent (gbent) functions defined on $\mathbb {F}_2^n$ with values in $\mathbb {Z}_{2^k}$ are regular, and show connections between the (generalized) Walsh spectrum of these functions and their components. Moreover we analyze generalized bent and semibent functions with values in $\mathbb {Z}_{16}$ in detail, extending earlier results on gbent functions with values in $\mathbb {Z}_4$ and $\mathbb {Z}_8$.

The rights of this work are transferred to the extent transferable according to title 17 §105 U.S.C.

A Appendix

Proof of Theorem 4 : Let $\mathbf{u}\in {\mathbb V}_n$. Eq. (2) for $k=4$ implies

$$\begin{aligned} \nonumber 16\sqrt{2}|{\mathcal H}_f^{(16)}(\mathbf{u})|^2&= (2+\sqrt{2+\sqrt{2}}) 4\sqrt{2}|{\mathcal H}_d^{(8)}(\mathbf{u})|^2+(2-\sqrt{2+\sqrt{2}}) 4\sqrt{2}|{\mathcal H}_{d+4a_1}^{(8)}(\mathbf{u})|^2\\&\qquad +8\sqrt{4-2\sqrt{2}}\ \mathfrak {I}\left( \overline{{\mathcal H}_d^{(8)}(\mathbf{u})}{\mathcal H}_{d+4a_1}^{(8)}(\mathbf{u})\right) . \end{aligned}$$

(4)

We denote by A, C, D, W the Walsh-Hadamard transforms ${\mathcal W}_{a_4}(\mathbf{u})$, ${\mathcal W}_{a_2\oplus a_4}(\mathbf{u})$, ${\mathcal W}_{a_3\oplus a_4}(\mathbf{u})$, ${\mathcal W}_{a_2\oplus a_3\oplus a_4}(\mathbf{u})$ (in that order). We denote by B, X, Y, Z the Walsh-Hadamard transforms ${\mathcal W}_{a_1\oplus a_4}(\mathbf{u})$, ${\mathcal W}_{a_1\oplus a_2\oplus a_4}(\mathbf{u})$, ${\mathcal W}_{a_1\oplus a_3\oplus a_4}(\mathbf{u})$, ${\mathcal W}_{a_1\oplus a_2\oplus a_3\oplus a_4}(\mathbf{u})$ (in that order). By [9, Lemma 17], we know that the generalized Walsh-Hadamard transform of any function in $\mathcal {GB}_n^8$, say d and $d+4a_1$ with $d = a_2+2a_3+2^2a_4$, is of the form

$$\begin{aligned} 4{\mathcal H}_d^{(8)}(\mathbf{u}) = \alpha _0 A+\alpha _1 C+\alpha _2 D+\alpha _3 W, \quad 4{\mathcal H}_{d+4a_1}^{(8)}(\mathbf{u}) = \alpha _0 B+\alpha _1 X+\alpha _2 Y+\alpha _3 Z, \end{aligned}$$

where $\alpha _0= 1+(1+\sqrt{2})i$, $\alpha _1=1+(1-\sqrt{2})i$, $\alpha _2=1+\sqrt{2}-i$, $\alpha _3=1-\sqrt{2}-i$, and moreover that (see also [9, Corollary 18]),

$$\begin{aligned} 4\sqrt{2}|{\mathcal H}_d^{(8)}(\mathbf{u})|^2&=A^2-C^2+2 C D+D^2-2 A W-W^2+\sqrt{2} (A^2+C^2+D^2+W^2)\\ \nonumber 4\sqrt{2}|{\mathcal H}_{d+4a_1}^{(8)}(\mathbf{u})|^2&=B^2-X^2+2 X Y+Y^2-2 B Z -Z^2+\sqrt{2} (B^2+X^2+Y^2+Z^2). \end{aligned}$$

(5)

Furthermore, with straightforward computations we get

$$\begin{aligned} \begin{aligned}&8\sqrt{4-2\sqrt{2}}\ \mathfrak {I}\left( \overline{{\mathcal H}_b^{(8)}(\mathbf{u})}{\mathcal H}_{b+4a_1}^{(8)}(\mathbf{u})\right) \\&=2\sqrt{2-\sqrt{2}}\, (B C + B D - A X - W X - A Y + W Y + C Z - D Z)\\&\quad +2\sqrt{4-2\sqrt{2}}\, (B D + W X - A Y - C Z). \end{aligned} \end{aligned}$$

(6)

Using (5) and (6) in Eq. (4) we obtain

$$\begin{aligned} \nonumber&16\sqrt{2}|{\mathcal H}_f^{(16)}(\mathbf{u})|^2\\ \nonumber&= 2 (A^2 + B^2 - C^2 + 2 C D + D^2 - 2 A W - W^2 - X^2 + 2 X Y + Y^2 - 2 B Z - Z^2)\\ \nonumber&\quad + 2 \sqrt{2} (A^2+B^2+C^2+D^2+W^2+X^2+Y^2+Z^2)\\&\quad +\sqrt{2-\sqrt{2}}\, (A^2 - B^2 + 2 B C + C^2 + D^2 + W^2 - 2 A X - 4 W X - X^2 \\ \nonumber&\qquad + 2 W Y - Y^2 + 4 C Z - 2 D Z - Z^2)\\ \nonumber&\quad +2\sqrt{2+\sqrt{2}}\, (A^2 - B^2 + B D + C D + D^2 - A W + W X - A Y - X Y\\ \nonumber&\qquad - Y^2 + B Z - C Z). \end{aligned}$$

(7)

If f is gbent in $\mathcal {GB}_n^{16}$, i.e., $|{\mathcal H}_f^{(16)}(\mathbf{u})|^2 = 2^n$, by the linear independence of $\{1,\sqrt{2},\sqrt{2-\sqrt{2}},\sqrt{2+\sqrt{2}}\}$ (as easily shown, the set forms a basis of ${\mathbb Q}(\sqrt{2},\sqrt{2-\sqrt{2}})$), we arrive at the following system of equations with solutions in $\mathbb {Z}$,

$$\begin{aligned} \begin{aligned}&A^2 + B^2 + C^2 + D^2 + W^2 + X^2 + Y^2 + Z^2=2^{n+3}\\&A^2 + B^2 - C^2 + 2 C D + D^2 - 2 A W - W^2 - X^2 + 2 X Y + Y^2 - 2 B Z - Z^2=0\\&A^2 - B^2 + 2 B C + C^2 + D^2 + W^2 - 2 A X - 4 W X - X^2 \\&\qquad \qquad \qquad \qquad \qquad + 2 W Y - Y^2 + 4 C Z - 2 D Z - Z^2=0\\&A^2 - B^2 + B D + C D + D^2 - A W + W X - A Y - X Y - Y^2 + B Z - C Z=0. \end{aligned} \end{aligned}$$

(8)

Let $2^t$ be the largest power of 2 which divides all, A, B, C, D, X, Y, Z and W, i.e., $A=2^t A_1$, etc., with at least one of the $A_1,B_1,\ldots$ being odd. First, if n is even and $t>\frac{n}{2}$, then $t=\frac{n}{2}+1$ only. Dividing by $2^{2t}$, the first equation of (8) becomes $A_1^2 + B_1^2 + C_1^2 + D_1^2 + W_1^2 + X_1^2 + Y_1^2 + Z_1^2=2$, which gives the solution $(\pm 1,\pm 1,0,0,0,0,0,0)$ (and permutations of these values). However, a simple computation reveals that none of these possibilities also satisfies the last three equations of (8). If n is odd and $t>\frac{n+1}{2}$, then $t=\frac{n+3}{2}$, but this implies that only one value out of $A,B,\ldots$ is nonzero. Again, that is impossible to satisfy the last three equations of (8). Assume now that $t<\frac{n}{2}$. The first equation of (8) becomes $A_1^2 + B_1^2 + C_1^2 + D_1^2 + W_1^2 + X_1^2 + Y_1^2 + Z_1^2=2^{n+3-2t}$, which is divisible by $2^5$ (when n is even, since $t\le \frac{n-2}{2}$), respectively $2^4$ (when n is odd, since $t\le \frac{n-1}{2}$). If n is even, this can only happen if $A_1,B_1,\ldots ,$ are all even, that is, $\equiv 0,2,4,6\pmod 8$, but that contradicts our assumption that t is the largest power of 2 dividing $A,B,\ldots$. If n is odd and $t\le \frac{n-3}{2}$, the previous argument would work, and if $t=\frac{n-1}{2}$, then $A_1^2 + B_1^2 + C_1^2 + D_1^2 + W_1^2 + X_1^2 + Y_1^2 + Z_1^2=16$. By considering every residues for $A_1,B_1,\ldots ,$ modulo 4 and imposing the condition that the 2nd, 3rd, 4th equations of our system also must be 0 modulo 16, we only get possibilities (0, 0, 2, 2, 0, 0, 2, 2), (0, 2, 0, 2, 0, 2, 0, 2), (0, 2, 2, 0, 0, 2, 2, 0), (2, 0, 0, 2, 2, 0, 0, 2), (2, 0, 2, 0, 2, 0, 2, 0), (2, 2, 0, 0, 2, 2, 0, 0) for $(A_1,B_1,\ldots )$ modulo 4, but that implies that all $A_1,B_1,\ldots$ are even, contradicting our assumption on t. This shows that the only possibility is $2^t = 2^{n/2}$ if n is even, and $2^t = 2^{(n+1)/2}$ if n is odd.

Thus, one needs to find integer solutions for the equation $A_1^2 + B_1^2 + C_1^2 + D_1^2 + {\mathcal W}_1^2 + X_1^2 + Y_1^2 + Z_1^2=8$, for n even, or $A_1^2 + B_1^2 + C_1^2 + D_1^2 + {\mathcal W}_1^2 + X_1^2 + Y_1^2 + Z_1^2=4$ for n odd, which also satisfy the last three equations in (8). Mathematica renders the following: if n is even, then $2^{-\frac{n}{2}}(A, C, D, W, B, X, Y, Z)$ (note the order) is one of

$$\begin{aligned} \nonumber&\pm (-1, -1, -1, -1, -1, -1, -1, -1),&\pm (-1, 1, -1, 1, -1, 1, -1, 1), \\ \nonumber&\pm (-1, -1, -1, -1, 1, 1, 1, 1),&\pm (-1, 1, -1, 1, 1, -1, 1, -1), \\&\pm (1, -1, -1, 1, -1, 1, 1, -1),&\pm (1, 1, -1, -1, -1, -1, 1, 1), \\ \nonumber&\pm (1, -1, -1, 1, 1, -1, -1, 1),&\pm (1, 1, -1, -1, 1, 1, -1, -1), \end{aligned}$$

(9)

and, if n is odd, then $2^{-\frac{n+1}{2}}(A, C, D, W, B, X, Y, Z)$ is one of

$$\begin{aligned}&\pm (-1, 1, 0, 0, -1, 1, 0, 0),&\pm (-1, 1, 0, 0, 1, -1, 0, 0),\\&\pm (0, 0, -1, 1, 0, 0, -1, 1),&\pm (0, 0, -1, 1, 0, 0, 1, -1), \\&\pm (0, 0, 1, 1, 0, 0, -1, -1),&\pm (0, 0, 1, 1, 0, 0, 1, 1), \\&\pm (1, 1, 0, 0, -1, -1, 0, 0),&\pm (1, 1, 0, 0, 1, 1, 0, 0). \end{aligned}$$

We see that in both cases, if f is gbent, then the conditions of the theorem are satisfied. The converse follows with straightforward calculations.    $\Box$

Proof of Theorem 5 : Assume that f is gsemibent in $\mathcal {GB}_n^{16}$ when n is odd, respectively generalized 2-plateaued when n is even. Then $|{\mathcal H}_f^{(16)}(\mathbf{u})|\in \{0,\pm 2^{(n+1)/2}\}$ for n odd, respectively, $|{\mathcal H}_f^{(16)}(\mathbf{u})|\in \{0,\pm 2^{(n+2)/2}\}$ for n even. Using the notations of Theorem 4, from Eq. (7), we immediately get $A=B=C=D=X=Y=W=Z=0$ if ${\mathcal H}_f^{(16)}(\mathbf{u})=0$. If $|{\mathcal H}_f^{(16)}(\mathbf{u})|=2^{(n+1)/2}$ (for n odd), respectively, $|{\mathcal H}_f^{(16)}(\mathbf{u})|=2^{(n+2)/2}$ (for n even), then (7) again yields the system of Eq. (8) with the one difference that in the first equation the power of 2 on the right side is $2^{n+4}$, respectively, $2^{n+5}$. With the same argument as in the proof of Theorem 4 we see that for such $\mathbf{u}$, $2^{-\frac{n+1}{2}} (A, C, D, W, B, X, Y, Z)$ (for n odd), respectively, $2^{-\frac{n+2}{2}} (A, C, D, W, B, X, Y, Z)$ (for n even) can only take the values from Eq. (9).

Straightforwardly, one confirms that the converse is also true, and the theorem is shown.    $\Box$